Prove the series (x^n)/(n!) is e^x

Problem: Prove the series (x^n)/(n!) is e^x

I'm not really sure where to start here. We have just learned power series in class, but not specifically Taylor series yet, so I can use any of those techniques yet. I know that the derivative of the series remains the same, but I don't know how to apply this to prove it is e^x.

I'm not really sure where to start here. We have just learned power series in class, but not specifically Taylor series yet, so I can use any of those techniques yet. I know that the derivative of the series remains the same, but I don't know how to apply this to prove it is e^x.

I think the solution is already in your message, but obviously you haven't yet studied this or perhaps you forgot it: there is one unique function which is infinitely differentiable everywhere, which equals 1 at zero and which equals its own derivative, and this is the exponential function ...
Perhaps there's another way to solve your problem, but I can't remember it right now.

Now we prove that (1) and (2) are equivalent. We start from the binomial expansion...

(3)

... where...

(4)

In order to well understand the next step we write the sequence of polynomials taking in evidence the powers of x from 0 to 3...

(5)

We now observe the coefficients of the 'by colums'. For and they are and they don't change by increasing . For the coefficient is and for it tends to . For the coefficient is and for it tends to . In general is so that for the coefficient of tends to . The conlusion is that...